text:
What is the stall speed for the Dassault Rafale?
Looking at this video of a Rafale escorting a high-speed train (TGV)

I was wondering what the stall speed for the Rafale is at close to sea level altitude.
Is this information generally public for fighter aircraft? I could conceive this would be classified.

Not an answer to your question but a piece of related information for readers : the commercial top speed of this train is about 300 km/h (video comment says it was going 320km/h).

Nice question! Note: with this train going to 320kph (198mph) I suppose it is far from stall speed

Lower than a train top speed, I'd guess ;-D

From Dassault's site: "Approach speed: less than 120 knots". If approach speed is 1.3 Vs, Vs is less than 100 kt (185 km/h). And btw this train has a speed record of 574 km/h (310 kt), a video of the record on YT. Stall speed has to be understood as speed where CLmax is reached in unaccelerated straight and level flight.

@mins to be more precise, this train class has a speed record of 574 km/h, but this was on a special project.

Your question is unclear. What stall speed are you asking for? Clean configuration? Landing configuration? Wings level or banked? Load factor of 1.0 or other than 1.0? Stall speed varies greatly. Conversely, AoA is constant for a given configuration.

@le_daim: Interesting article. It's generally well understood the wheels were larger and the track is descending, as it was the case near Vendôme for the previous record. By the way the plane mentioned in the article is a Corvette with a stall speed of 91 kt (168 km/h).

I guess the ideal answer would have a table of stall speed by configuration/weight/height ? At minima, it would be "what is the lowest speed achievable in level flight for any configuration and realistic amounts of fuel (1h or more left).

80 knots for an aggressive pilot, as said here.
It is 148 km/h, a speed of fast but rather typical long range passenger train. No new technology is required for the train, this plane could easily follow a Pacific steam locomotive (top speed 203 km/h).

It must be said that in airshows the airplane flies almost empty - no munitions, no external loads (even the missiles can be dummies) and little fuel in the tanks. With only 10t empty weight, it reaches those 80 knots at half the wing loading of a fully loaded plane. This would make the "real" stall speed closer to 113 knots.


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How big was the F-14 fleet compared to other US carrier-based aircraft in 1981?
Back in the 80's I was away jesting about the F-14 are here to protect the carriers and the carriers to deploy F-14.
I know there are a number of other aircraft here for other purposes but I really don't know the numbers.
So for 1981 what's the proportion of F-14 to other aircrafts and also de percentuals in roles (defence of the fleet, attack, surveillance, etc).
In other words: for US aircraft carriers of that period, what was the aircraft fleet composition?

Which ship or which class of ships are we talking about here?

@user3528438 All US aircraft carriers ships in service in 1981? https://en.wikipedia.org/wiki/List_of_aircraft_carriers_of_the_United_States_Navy

This might be useful: https://en.wikipedia.org/wiki/Carrier_air_wing#Cold_War_(1974%E2%80%931990)_and_the_1983_Invasion_of_Grenada

F-14s were aboard all large deck carriers, except for the Midway class ships and the training carrier Lexington.

Are you asking about the Embarked Airwing, or about F-14's?

@KorvinStarmast Embarked airwing and F-14 role and numbers comparation with the other aircraft here

Your bottom line question and your title question don't match; why don't you rephrase your title to match your bottom line question?

During the plane’s heyday, the US Navy maintained 26 active duty F-14 squadrons for the 12 carriers certified for F-14 operations, 4 Test and Evaluation squadrons, 2 Fleet Replacement squadrons and 4 Naval Reserve squadrons.  632 F-14s were delivered to the US Navy between 1973 and 1994.  Each Forrestal-, Kitty Hawk-, Enterprise-, Kennedy-, and Nimitz-Class carrier had two squadrons of F-14s aboard, with 12 aircraft per squadron.  Depending on the cruise air wings aboard, there were approx 85 aircraft aboard ship.  This sets the F-14s comprising about 28% of the air wing’s airframes.
The F-14 was used to provide air superiority in the airspace around the battlegroup as well as escorting strike packages and providing a first line of defense against long range enemy interceptors and cruise missiles.  Later upgrades as well as the F-14D variant allow for precision strike capability against surface targets as well.  The genesis of the AWG-9/AIM-54 weapons system was as a defense against TU-22s carrying Kitchen cruise missiles which had a very real chance of sinking a carrier if the missile found its mark.  A secondary role was given to the F-14 in the early 80s as the primary tactical reconnaissance aircraft for the fleet, replacing the aging RA-5C Vigilante and RF-8 Crusader aircraft by means of a removable Tactical Aerial Reconnisance Pod System (TARPS) which mounted on the center body pylons for the AIM-54 missile.  Later systems included ROVER pods for precision close air support.

Not doubting you, but some sources would turn an good answer into a great one. Also, the "TARPS which ovule Mount"... care to fix a typo or two there?

This also doesn't show how these numbers compared to the total naval air fleet.

https://en.m.wikipedia.org/wiki/Grumman_F-14_Tomcat

Also Robinson, Anthony, US Air Power.

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text:
What type is this Syrian light attack aircraft?
I saw the following 2012 photo of a light ground attack jet used by the Syrian regime in an attack on Tel Rafat:

I do not recognize the type, however, and it does not appear to be any of the regular ex-Soviet types normally flown by the Syrian air force (MiG-29, MiG-25, MiG-23, MiG-21, Su-24, Su-22). Does anybody know what it is?

Possibly an Aero L-39 Albatros: See images here

Its a Syrian Aero L-39 Albatros.
You can see the same image (just a bit sooner on the second missle lauch) when searching Google for Syrian L-39 (second result).

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text:
Why do "angels" mean MSL, not AGL?
In military aviation, crews will sometimes refer to "angels" meaning MSL altitude. However, one might guess that it is a pronunciation of AGL (AnGeL), above ground level. Did it originally mean AGL and then at some point it changed meaning to MSL?

It's never meant above ground. For one thing a pilot doesn't always know how far above ground they are as that changes all the time. AGL is useless information to broadcast as it doesn't give a common form of reference. I can't tell if I'm at risk of collision if someone says they are 10000ft agl.

I highly doubt that the abbreviation "AGL" was in widespread use at the time the "angels" term first started being used.

It was originally Royal Air Force slang for altitude in thousands and not a way of pronouncing AGL.
Up in the air, where the angels fly.
https://en.wikipedia.org/wiki/RAF_slang
Angels 10 meant 10,000 feet.  Cherubs were hundreds of feet.

The question is still unanswered: Was "Angels" always MSL or was it once AGL?

@MikeSowsun I'm not sure.  I've looked for references, but I doubt there are any easy to find ones.  I believe that regional QNH was used. I doubt that the difference in AGL and MSL would be very important anyway since the accuracy of the RADARs or spotters used would not be less than the difference.

My main point is that angels was not some way of pronouncing AGL.

I agree, it probably had nothing to do with AGL and was always MSL.

@Simon if that is your main point, I strongly recommend that you edit your answer because I see no indication of that being the main idea.  My main take away from your answer is that it means altitude in thousands of feet but nothing about a reference datum.

@MortensenAviation Why.  Is it not clear?  I quote, "it was originally Royal Air Force slang for altitude in thousands of feet".  I believe that this quite clearly shows that it is not some way of pronouncing AGL.  What is missing?  From the question - "Did it originally mean AGL".  My answer is factually correct and answers, at least partially, the OP question.

Also worth noting that cherubs is height agl, I guess because height is more useful than altitude when you get to below 1000 ft.

@DanHulme Hi Dan.  Do you have a reference for this?  I've been trying to find some.

No, it always meant altitude and altitude means above mean sea level since it is distinguished:

Back in 1930s and 1940s when the term was already in use, there was no way to find height above ground level at all. Even now the radar altimeters only measure height up to few thousand feet.
The term is mainly used to describe position of flights over the battle area, for which you need something that will immediately tell you whether the other flight is above or below you. Altitude is such thing—higher number means above, lower number means below. Height wouldn't work as a flight over mountains can have lower height AGL and still be above you.

QFE and QNH were (and still are) used to measure height with pressure altimeters.  QFE gave height above ground.  "Angels" was not measured by aircraft, it was the reported height of attacking aircraft given by the "chain home" RADAR stations or ground observers.

@Simon, well, QFE gives height above airport (field elevation), which is different thing from height AGL. QNH, of course, gives altitude. Thousands of feet is low enough precision that the difference between QNH and QNE does not really matter.

It was my understanding that all a radar altimeter can tell you is when you are about to crash or land.

@HowardMiller, not exactly. Radar altimeter measures height above ground, accurately to few feet, but only up to few thousand feet. The thing that tells you when you are about to crash or land is (E)GPWS, which uses the radar altimeter output, but in the newer versions also GPS altitude and position and a map so it can tell you when there is a rising ground ahead of you—the radar altimeter only measures directly down. But all these things are much newer than the term “angels” anyway.

"Angels" is probably still around because it sounds cool in movie dialog. Like "Breaking atmo".

Close to the ground, altitude AGL matters far more than altitude MSL, since it is the ground that you're trying to land on, or avoid running into. At higher altitudes, however, altitude AGL matters far less, and is much harder to determine, and thus altitude MSL is used as the reference. That is also what is read on the barometric altimeter.
Thus, for an aircraft looking to join up with (or avoid hitting) another aircraft, giving the altitude in the same reference (MSL, not AGL) that can be readily read on the instruments is most useful.
Besides, if that aircraft is over a mountain and I'm over a valley, we might be at the same altitude even though he is at 4000' AGL (above the ground directly beneath him) while I'm at 6000' AGL (above the ground beneath me).
One might ask, why not set QFE and reference everything to the "ground level" at some particular airport, rather than referencing everything to sea level.  You could do that, AS LONG AS everybody has the same QFE for the day.  If you're doing all your flying in a theater where all the airfields are at about the same altitude, this probably wouldn't cause much problem, but if I have QFE set for the field I took off from (elevation 10' MSL), that setting won't show me what I want to see if I end up landing at another field at say 500' MSL!
Thus, the fairly common choice to set altimeters to read height above sea level, and thus reference calls of other aircraft altitude to that same standard.

"Angels" is flight level in thousands of feet based on the common barometric pressure of (I think) 1013.25mbars (29.92mmHg). Angels is not typically used when you have a QFE or QNH to reference and is usually used above a certain altitude only and when away from all other references. For example a transatlantic flight once over the ocean could use MSL, but in a transmission to other aircraft that are currently using QFE or QNH over land, this reading is not right for them to reference for anti collision. To that end, once you are in an air corridor with other aircraft, and I think usually above about 18,000ft, the altimeter is set to a common value of, like I said, I think, 1013.25mbars and then FL is used so all are on a common altitude reference.

On another note AGL primarily stands for Aeronautical Ground Lighting as in runway/airfield lighting systems that pilots rely upon to safely land aircraft including PAPI lights.

No, that's not what AGL means in this context. VTD as not an answer to the stated question.

AGL is primarily used to denote Above Ground Level. In my many years at the FAA I never heard it used to reference lighting.

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What is the minimum turning radius of an SR-71?
What is the minimum turning radius of an SR-71 at Mach 3.2 and an altitude of 80,000 feet?
I have heard that if an SR-71 were to cross the Pacific coast over San Francisco and pull a hard turn to the right, it would return over Seattle.
I'm interested in real-world performance, taking into account things like inlet unstart (failure to capture the engine inlet shockwave in the intake) from too sharp an angle of attack.

Please use the equations of this answer. The numbers might be different, but the physics are the same.
EDIT: Thanks to D_S for providing the link to the manual.
When flown with the maximum allowable load factor of 1.5 g at 80.000 ft (48° bank), the turn radius at Mach 3.2 (equivalent to v = 953.3 m/s in 80.000 ft) will be 83.5 km. To be more precise, you will need to add the effects of earth rotation, but for now I leave this away. As you can see, the turn will still need 163 km or 103.7 miles, but not the distance from San Francisco to Seattle which is more than 6 times bigger.
To turn this around: A circle at Mach 3.2 which has a diameter of 1092 km requires a bank angle of 9.6°. That can hardly be called a turn.

The equations are nice from a theoretical standpoint, but you need to pick your assumptions when using them.  The SR-71 is so far outside of "normal" for airplanes that I don't know what assumptions are reasonable.

@Mark: I refined my assumptions, but the SR-71 was not so unusual that it would violate the laws of motion. Those equations are valid for all airplanes.

The equations are valid. But the maximum possible load factor is an argument we need to know. The load factor not only needs to be allowed, but also achievable at cruise altitude and speed.

@Jan Hudec: At Mach 3.2 the possible lift is out of the question - increase angle of attack, and the lift will be there. The main problem is the drag increase which could be more than the engines can compensate (and the efficiency of the intake at the different angle of attack). 1.5 g should not be an issue, however, if the airplane was flying level with some margin before. All it needs is half empty fuel tanks, and there is plenty of margin already.

@Articuno: Because I use the formulas, and he some general figure. Besides, we are not so far apart. 80 NM are 92 miles, not so far off the 103.7 I have as a result. His result is valid for Mach 3.0 (the book page mentioned in the comments), my result is for Mach 3.2, as asked by Mark. Turning any tighter than that circle with 103.7 miles diameter at Mach 3.2 would violate the aircraft limits, though.

@Articuno: More than that. The diameter grows with the square of the speed (please look up the formulas in the first link in my answer!). His 80 NM at Mach 3.0 translate to 91 NM at Mach 3.2 or 104.7 miles. This leaves just a rounding error between both results. My answer is both physically and legally correct, and actually answers Mark's question.

@Articuno: In a comment D_S linked to Figure 1-9 in a book. If you look up the link, it gives you the conditions for his number. Yes, that is StackExchange: Not always goes the highest score to the correct answer. We are all humans, after all.

@Articuno look at the temperature mentioned above the chart and then check the definition of standard atmosphere: you'll find that is the temperature you have above 36000ft till 66000. 80000 is slightly "warmer" but not much.

@Articuno: Density is not so important. What counts is true air speed, and with the temperature we can convert Mach to TAS. I also used the Standard Atmosphere (from 1963) for my result, so both the chart and my result yield the same radii. The chart stops at 44° bank with a radius of 51 NM; if you use the equations in my link, at 44° the radius will be 15% bigger than in my result. Close enough!

@Articuno: The bank angle is the arccos of the inverse of the load factor. For 1.5g, the inverse is 0.6667, and the arccos of that is 48.1597 degree. If the pilots fly the turn without sideslip, this is the bank angle that produces a load factor of 1.5, which is the limit for the SR-71 above Mach 2.6 (see page 5-8 of the manual).

The Turn Radius of the SR-71 would depend on its speed. The faster it went  the wider its turn radius was.
The SR-71 had a minimum turning radius at altitude of about 80 nautical miles (NM) . It was not an airframe limitation but a matter of wing area. At 80,000ft, the air is too thin and the wings too small to allow for much lift to turn with.
At a turn radius of 80 NM, the SR-71 would cover about 145 miles, taking about 4 minutes in the process of making a 180 degree turn.
Details are provided in the SR-71 flight manual handbook, which is now declassified:

Check Figure 1-9 http://books.google.co.in/books?id=voPlAwAAQBAJ&pg=SA1-PA11&lpg=SA1-PA11&dq=turning+radius+at+mach+3.2&source=bl&ots=dr5oLR4Tho&sig=TGRKTrot7X8ucsCxnRuJ_8xK5t4&hl=en&sa=X&ei=nZDiU6fBHMOKuASEzYDoBQ&ved=0CFAQ6AEwBw#v=onepage&q=turn&f=false

Yes , that's where the answer is . And No , it cannot be in my answer because I do not have permission to re-produce the figure.

@D_S I can't imagine Lockheed Martin would object to reproducing this under the Fair Use doctrine: this would qualify as an "educational purpose", only a small portion of the overall work is being reproduced, and as the document is a (declassified) government work freely available to the public there is no impact on the market for / value of the work. If they take exception to the image being included here I'll gladly take the hit :)

The top answer states: "At a turn radius of 80 NM, the SR-71 would cover about 145 miles, taking about 4 minutes in the process of making a 180 degree turn." This should raise the reader's eyebrow. If the turn radius is 80nm, then after completing a 180 degree turn, the plane would be laterally offset by 160nm (184 statute miles) from its previous track. Its actual ground track would cover 251 nm (almost 290 statute miles) in that 180 degree turn.

I worked one in about 1989 that was flying eastbound toward  Rapid City, and started his turn back toward Edwards. He rolled out over GLL VORTAC, so about 200 nautical miles south of where he started. He was at roughly 2400 knots (He slowed down for the turn). Altitude was above FL600, but we did not get information on actual altitudes.

Considering the current flight status of the SR-71 (retired) it's maximum speed will depend on the tug moving it and the turn radius is probably around 30-40 meters depending on how far the nosewheel pivots.

I would like to see a tug that can pull an SR-71 at Mach 3.2 and 80,000 feet.


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Are composite materials, which are being widely used in newer aircraft, better at handling bird strikes?
Do current composite materials handle bird strikes better than their predecessors? Or they are not part of forward-facing components/areas?
For example, the Boeing Dreamliner or the Airbus A350-XWB.

Composite rotor blades use a nickel (extremely wear-resistant) strip on the leading edge to reduce erosion from blown dust/sand.

In general, the composite materials are poor compared to metals in impact resistance (in general, composites are not good in resisting compressive loads). This is because the metals and composites respond to impact in different ways.
In case of metals , most of the impact energy is absorbed by the metals through plastic deformation. As the plastic region is pretty large compared to the elastic region, this means that they have a good impact resistance.

Source: www.spaceflight.esa.int
However, the (fiber reinforced) composites undergo little plastic deformation. The response of the composites in case of impact depends on a number of characters like the speed of impact, fiber orientation etc. In case of low intensity impact, elastic deformations may appear near the impact area. A high speed impact, however, causes deteriorations of the material, which includes separation of the fibers from the matrix, matrix cracking and fiber
breaking.

Comparison of stress- strain curves of composites and metals. Image from Circular and Square Concrete Columns Externally Confined by CFRP Composite: Experimental Investigation and Effective Strength Models by Riad Benzaid et. al.
The absorbed energy consequent to the impact depends, among others parameters, on the fiber – resins link resistance. If this link is strong, a continuous crack may spread along the material. In the case of a weak link, the generated crack may have an irregular form, leading to a rapid separation of the fibers from the matrix and to a considerable absorption energy. However, the adhesion between the fibers and the matrix shouldn't be too weak as a low shear resistance also influences the impact behavior in  a negative manner.
The primary mechanism is that the fibers are subjected to bending, which breaks them, consequent to which the load is transferred to the resin base, which usually cracks. The the load is transferred to the other fibers, which get damaged.
There are some ways to make the composites better impact resistant, like using materials like Kevlar, which has higher impact resistance. These tough materials have higher failure loads,which helps in absorbing the impact load and transferring to to the resin matrix (which carries the compressive load).

Impact strength of various composites, from The Impact Behaviour of Composite Materials by Chircor Mihael et. al.
The composites can also be tailored to have a better impact resistance. Also, the places where the impacts are expected, like the leading edges can be made from metals. Interestingly, both A350XWB and Boeing 787 use metals in the wing leading edges, though I'm not sure how much of this is attributable to impact resistance.

Source: www.1001crash.com

On the other hand, Kevlar and kin are considerably harder than glass or carbon fiber to incorporate successfully into composite aircraft structure.

Out of the box, no. They are worse at handling birdstrikes. But with some thought they can be made much better.
All materials have an elastic range of deformation, and if that is not exceeded, they will assume their original shape without damage. Composites fail if this range is exceeded. First the fibers with the highest elongation fail, which enables the other fibers to take more load, but at some point too few fibers remain, and the part fails catastrophically. This is audible: The first small failures cause a faint crackling, and the eventual failure a loud bang.
Metals add a plastic deformation before ultimate failure, and this enables them to absorb a lot of energy in the process. Since energy is force times distance, the small elastic deformation will not absorb much energy. However, the plastic deformation is magnitudes bigger and makes metals the much better energy absorbers. Even a deformed structure can take a sizable fraction of the load it took before deformation. A birdstrike will cause a dent, but the rest of the structure will not see much of the load since the dent-forming could absorb a lot of energy. However, the changed contour can provoke changed aerodynamic loads, so new failure modes might result.
Impact loads on composites have a second disadvantage: The impact creates shockwaves which travel through the structure and cause delamination (separation between fibers and resin matrix) when they hit a boundary. Therefore, the biggest damage occurs on the opposite side of the impact. A birdstrike might not leave a visible dent on the outside, but can shatter the material on the opposite side such that the structure will break at a fraction of the load it would have carried without the bird strike.
Therefore, some composite wings use metallic leading edges to protect the composite structure from birdstrikes. But also the composite structure itself can be improved: If very tough fibers like Kevlar or Spectra are added, the impact energy will not break those fibers, and the load will be distributed over much more of the resin matrix. This will delay delamination, and when it happens, it will be spread over a bigger area and cause the fibers to break at multiple locations. The energy needed for the widespread delamination and the multiple fiber failures will increase the energy absorption dramatically. This structure must be designed as a protective armor, so the load-carrying structure below is not affected. Proper crashworthy design with composites allows to make the structure even tougher than a comparable metallic structure.

The important difference between the energy absorbed in elastic and plastic deformation is that plastic deformation is non-reversible, and the energy is retained within the material permanently as locked-in internal stresses. The energy stored in elastic deformation is only temporary, and is released when the structure returns to its original shape. This often causes vibrations to be transmitted through the structure which can take a relatively long time to die away, and may cause collateral damage at places a long way from the original impact site as they do so.

May be it's better to answer the question the other way around. There are regulations which need to be followed. This also includes bird-strikes. In order to get certified all the requirements have to be accomplished.
The reason new materials are used is because they feature desirable attributes (they are for example lighter or maybe cheaper) while achieving the requested properties.
It is not only a specific property of the material which makes it sustaining a bird strike but also the way the whole part was designed by the engineers. Depending on the material this design process might be different and might contain certain challenges. However, the final part will only be as strong as the regulations require. This will be true for aluminium, titanium or composite materials.


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What is the stall AoA for the space shuttle?
Once flying as (fast) glider in order to land, what was the angle of attack it would have stalled? I'm really interested in the angle of attack for such wings and not the speed.

@qqjkztd Although they may have good knowledge about all lots of things concerning the STS, especially all what makes it a rocket (SRB, tanks, engines, life support, reentry phase,...), I doubt they are concerned about AoA (which is a really specific aviation related thing). I may be wrong, I'll try asking there if I have no answer here within a reasonable time.

Related: https://aviation.stackexchange.com/a/23889/1379

Just google "stalling AOA for Delta Wings" and you'll find what you need.  They can operate to extreme AOAs without a break in the normal sense as the vortices made by the LE angle mitigate or prevent the flow separation it would normally suffer from.  They just get draggier and draggier.

It depends on what you mean by "stall". In some definitions (eg. from the perspective of a pilot) delta wings may not have any stall angle. They just change glide angle vs AOA without ever losing control

@ManuH Some of SESE members absolutely do have relevant documentation.

@slebetman the definition given in how it flies, which seems to be a consensual one.

As far as I can tell, NASA has never given a straight-forward answer to this question. The theoretical stall AoA is likely in the 33-40° range (see reason for that guess below). However, at operational airspeeds, the orbiter becomes aerodynamically unstable at a lower angle of attack than the stall angle. Therefore, the stall angle never mattered for shuttle operations.
Just to be clear, this answer is only referring to the very late phases of shuttle flight (TAEM and approach and landing). The 40° of early entry might be beyond the wings' stall angle, but it doesn't really matter since dynamic pressure is low and the shuttle is on more of a ballistic trajectory.
What does NASA say?
From an old NASA answer to this question:

The question: "What is the Space Shuttle stall speed?" is a bit of a
  trick question because, as a delta wing aircraft, the Orbiter does not
  have a clearly defined stall point.
...
We have determined the minimum flyable speed based on the longitudinal
  stability characteristics.  We start by evaluating the maximum angle of
  attack, alpha, for which the Orbiter is aerodynamically stable in the
  pitch axis.  This is the angle of attack at which the pitching moment
  partial derivative with respect to alpha changes from negative (stable)
  to positive (unstable).  This angle of attack is identified as the
  maximum usable angle of attack and it varies by Mach number.

From The Aerodynamic Challenges of the Design and Development of the Space Shuttle Orbiter document:

The low-speed longitudinal characteristics shown in figure 29 demonstrate stall-free characteristics over operating flight conditions.

The diagram on the left shows that the lift coefficient continues to increase through at least ≈25° angle of attack (meaning the stall angle must be greater than 25°). δe = elevon deflection.
The only* reference to stalling in the Shuttle Crew Operations Manual is this diagram which depicts a stall speed (not angle). The NASA answer I referenced before gives more context to this graph.

Shuttle Crew Operations Manual - Section 4.9-1
*There are a couple references to elevon stalls in the Operations Manual, but I'm not counting those.
Why does the stall angle not matter?
As we've already seen, the shuttle would exceed aerodynamic/structural limits before reaching the stall angle of attack at most airspeeds. Additionally, there are a couple other reasons for not worrying much about the stall angle.
Guidance always aimed to keep the orbiter on the "front side" of the L/D curve. This means that guidance would never command an angle of attack higher than required for "best glide". If it needed to dump energy, it would do so by deploying the speed brake and/or pitching nose down. It would never dump energy by pitching up (except during the flare). If the orbiter was low on energy, it would pitch for best glide. This means airspeed would always stay high and in the range where other factors are more limiting than the stall angle. Pitch-up maneuvers would be limited by dynamic pressure and Nz (normal acceleration — g's).
The only time the shuttle fell into the "back side" of the L/D curve was during the flare and landing. Incidentally, touchdown speed appears to be roughly the speed where the orbiter is finally slow enough that it might actually stall before becoming unstable for other reasons (see "VA" in the graph above - this is "maneuvering speed", just like in airplanes). However, at touchdown, the limiting angle is the tailscrape angle (15°), which is much lower than the stall angle.

Shuttle Crew Operations Manual - Section 7.4-17
What's the real answer?
Although not operationally significant, I think the specific question you're trying to ask is: what AoA produces the maximum lift coefficient? Generally speaking, that's considered the stall angle for most airfoils, since increasing AoA beyond that point results in decreased lift.
While I don't have a lift coefficient graph for the shuttle which goes beyond 25°, this paper discusses the results from studies on various delta and double-delta wing designs.

None of the studied designs have the shuttle's exact 81/45 sweep angles. However, given that they all have a maximum lift coefficient somewhere in the 33-40° AoA range, I suspect the shuttle was similar. I can't prove that unless someone finds the appropriate data. I'm certainly not an expert on delta wings, so my interpretation could also be wrong.

Great, well sourced answer.

@OrganicMarble coming from you, that means a lot! :)

The definition of stall given in how it flies is exactly the one you use in the last section. i'm tempted to consider it as a valid and consensual definition of stall.

I would expect that the shuttle would lose directional stability long before reaching its maximum AoA.


text:
What triggers the PTU (Power Transfer Unit)?
As highlighted by this answer and by this page the PTU (power transfer unit) is an hydraulic driven pump. It gets energy from one hydraulic system and uses it to pressurize another hydraulic system.
What triggers the functioning of this pump? Is it automatic or manual? Is this pump available even if a leak is detected in the hydraulic system it tries to restore pressurisation to?

(A320 PTU (Vickers), source)
The Power Transfer Unit is well known on Airbus due to its barking-like sound (refresher: Video of a PTU in action). The sound comes from a pump which creates a 3,000 psi pressure being quickly switched ON and OFF.

(Source: Youtube)
The PTU is a safety system to ensure hydraulic pressure is available, specially at takeoff, even if one or two hydraulic circuits have failed.
The PTU can restore pressure in a deficient circuit without actually transferring fluid from the working circuit to the deficient one. This prevents throwing hydraulic fluid from the working circuit overboard if the deficient circuit is leaking.
The PTU is always activated, but is actually triggered based on a pressure differential.
When is it activated?
On Airbus A320 series, the PTU is actually the second source of pressure for the yellow and green hydraulic systems, after the EDP. It's always ON, ready for the pump activation if there is a pressure drop on the yellow or green hydraulic system and the differential pressure is more than 500 psi.

Airbus A318, A319, A320 and A321 aeroplanes are equipped with three
  hydraulic systems and they can be safely controlled with only one
  system operative. A bi-directional PTU enables the yellow system to
  pressurize the green system, and vice versa, without the transfer of
  fluid from one system to the other.
In flight, the PTU operates
  automatically if a pressure drop in one of the systems is detected.
  The PTU ensures optimal hydraulic system availability for take-off.
(Source: EASA safety bulletin)

Automatic inhibition below 1,500 ft in case of leakage
The PTU cannot restore the correct pressure in a circuit with a permanent leak. In this case, the PTU, and possibly the working hydraulic system, eventually overheat. The crew receives a failure amber caution on the ECAM display and, up to recently, had to switch OFF (inhibit) the PTU manually to prevent damage.

In case of loss of hydraulic fluid of the yellow or green system, the PTU
  cannot pressurize the failed system and must be switched OFF as
  required by ECAM procedure to avoid a PTU overheat which may occur two
  minutes later.
(Source: EASA safety bulletin)

A320 PTU switch (source: Baltic Aviation Academy which, by the way, has plenty of quality training videos)
To avoid distraction from non-essential warnings during takeoff and landing, the ECAM message is not displayed when airborne below 1,500 ft height. There were several incidents (e.g. G-EZDM) where overheating became harmful before the amber ECAM message appeared at 1,500 ft, causing the temporary loss of a second system. Airbus added logic for automatic PTU inhibition when the aircraft is airborne, but still below 1,500 ft.
The previous quotes are from an EASA bulletin related to this modification (also detailed in Safety First, the Airbus safety magazine), with a failure scenario that had to be prevented:

Principle
A PTU is connected to two independent hydraulic systems: The first to drive an hydraulic motor (or turbine) mechanically connected to a pump. The second hydraulic system is pressurized by the pump.

PTU principle, source
PTU can be reversible (each hydraulic system can drive the motor or be pressurized by the pump) or not (the same system always drive the motor).

(Actual PTU schematics, source)

Airbus A320 (also A319) has three hydraulic systems — green, yellow and blue. Green is driven by engine 1, while the yellow is driven by engine 2 (there are electric pumps for yellow and blue, but that's beside the point). The PTU is a bidirectional unit which enables the yellow system to pressurize the green and vice versa.
The PTU runs automatically when the differential pressure between the green and yellow systems is more than 500 psi (normally, the system pressure is 3000 psi). The PTU (hydraulic) control/indicators in the overhead panel is shown in the following image:

Image from A320 hydraulic manual
The PTU is inhibited during the first engine start and self tested when the second engine starts. During flight, the flight crew receive warnings for the following situations:

Overheat in green/yellow reservoir.
Low air pressure in green/yellow reservoir.
Low fluid level in green/yellow reservoir (due to leak).

In these cases, the PTU will be inhibited automatically and will show a fault if it is still in AUTO mode, while the crew can take appropriate action.
References: A320 hydraulic Manual

IIRC the inhibition was added (or extended, not sure) as a modification relatively recently due to number of incidents where it overheated before the crew made it to the checklist item that orders turning it off.

Did you mean to say air pressure? If so why?

PTU Logic requires:
BOTH ENG MASTERS ON POSITION
DIFF PRESSURE > 500 PSI BETWEEN GREEN AND YELLOW HYDRAULIC SYSTEM
NOSE GEAR COMPRESSED
PTU SIWTCH AUTO POSITION
PARKING BRAKE SET TO OFF
PTU Inhibited below 1500 and airborne, otherwise Operating Engineering Bulletin for the aircraft with instruction to switch off when encountering hydraulic overheat or leakage.
Also dúring engine start, the PTU is tested by the Flight Warning Computers --> barking sound.
Cheers

The A320 has a yellow hydraulic system pressure indication on the ECAM which shows xx =even the press by edp or by yellow elect pump = the press txd replaced -the power to txd measured ok 28 v=xx indication on yellow sys is always with press applied or without.

Your answer is really hard to follow. You may rewrote it so that it is easier to read.


text:
How is airbrake function performed on the Dassault Rafale?
At the end of airshow display (just before landing), the dassault rafale seems to slow down very quickly as it reach the top a final loop. It seems as if the pilot apply full airbrakes, but it seems there is no special flight control surface deployed to perform this function. How is this feature performed? Is it a special configuration of the split elevons? A combination of elevons and canard?

I've watched the video and cannot see this slow down.  Please could you edit your question to include the exact time?  I assume you mean about 7.45 but I don't see any sudden slow down.  Just a half roll to the inverted , a bit like an Immelman, and a loss of energy vertically as he reaches the apex.

@simon the link leads to the specific time in the video. This minor part of the question is just an illustration. The important part of the question is "how is airbrake function performed [given] there is no [dedicated] flight control surface?"

The Dassault Rafale has no dedicated airbrakes. By having two means of pitch control, namely wing elevons and canards, in combination with relaxed stability and an advanced FCS, it can move both into extreme deflection angles without sacrificing pitch control. This site states:

When landing, both canards and trailling-edge control surfaces can be
  used for braking, and Rafale may be able to use canards for braking
  even while in flight.

Also, having little load to carry helps. The display aircraft was most likely flying close to its minimum wing loading (which is just 276 kg/m²), and by cutting back thrust could employ its regular aerodynamic drag for braking.


text:
What is the Space Shuttle's command law?
The Space Shuttle was fly-by-wire. In today's fly-by-wire systems, the command law commands either the load factor and the roll rate (e.g., A320's normal law), or the surface control deflection (e.g., A320's direct law).
What was the case for the Space Shuttle?

For future reference, the term "command law" is specific to Airbus aircraft. Most other manufacturers of fly-by-wire aircraft use the term "mode" to refer to the same thing. You'll probably get better answers using the more common term.


text:
What is the highest operational ceiling for an air-breathing jet engine?
An air-breathing jet engine can only operate at relatively low altitude, where the atmosphere is dense enough. However, some jet-powered aircraft can operate at an altitude higher than 20 Km (e.g., the SR-71). For really high altitude, aircraft are equipped with rocket engines (e.g., the X-15)
What is the highest operational altitude for any air-breathing jet-powered aircraft?

I think somewhere on Wikipedia has a page which lists the highest aircrafts by category: rocket engines / air breathing engines / blimps etc.

From what I found: MiG-25 37,650 metres (123,520 ft)

@kevin I did find a page about altitude records but none about operational ceiling

A Scramjet engine, which is still technically an Air breathing engine, will provide the highest possible service ceiling.
Theoretically, they can operate at up to 250,000 ft (75 km). For comparison, "space" begins at around 330,000 ft (100 km).
More technical info about Scramjet engines can be found here.

You should had more references for further readings and to sustain your claim of theoretical ceiling for scramjet. E.g. a research paper from the nasa is quite more authoritative than a wikipedia link and provide more insight

@ManuH Maybe this will give you some more info: http://upcommons.upc.edu/bitstream/handle/2099.1/20295/Technical%20Report.pdf

@JimPP you really should edit the question. such a link is better fitted in the answer than in its comments.

@ManuH I will, thanks!

Where the atmosphere "ends" and space "begins" is a matter of definition. The atmosphere's sea level composition (78% N2 21% O2 1% Ar) ends at the turbo-/mesopause around 53 mi (85 km) altitude. Hydrogen and helium mostly prevail above 550 mi (880 km). The ozone layer is thickest around 15 mi (24 km). Water boils at body temperature at 12 mi (19 km).


text:
How often is the name painted on fighters?
This question is restricted to US Air Force fighters, as I don't know if other air forces paint the pilot's name on the fighter.
This question pops into my mind because of 2 reasons:

this comment reminds me a fact I saw on a comic strip long time ago: the name of the pilot is written on the fighter cockpit, at least in the US Air Force and perhaps in the US navy.
in this video (smarter everyday, episode235) there is one scheduled flight with the videographer. For this single flight they took the time to write the videographer's name on the fighter (see at 6min17s).

How often is this name (re)painted?

is it updated for every scheduled flight if needed?
is it updated for a single flight flown by another pilot of the same squadron?
is it updated when the fighter is operating on an external theater?
if no update is needed, is it repainted more often than the rest of the airframe?
does this answer varies from squadrons to squadrons?

Bonus question: is it actually painted (I may have made a false assumption, in which case I should substitute "painted" by another more suited word in the whole question)?

Reminds me of a scene from some movie or TV episode I saw a long time ago where the WW2 fighter pilot scores a victory in combat and then right there in mid-flight he slides back the canopy and reaches out and slaps a decal on the side of the fuselage representing his latest "kill"-- yeah, right!

@quietflyer was that a TV episode of Donald Duck or Mickey Mouse? Sounds like one of the WWII propaganda films they produced...

@FreeMan  -- no, it was a human actor, but I don't recall whether the whole thing was meant to be a ridiculous slapstick spoof, or not.

I cannot answer for the USAF, but Navy and Marine Corps squadrons have also done this for years.  In addition to the pilot, if there is more than one seat the Weapon System Officers name(s) will also be painted, in addition to the Plane Captain.  It is an honor to have your name on a plane, and it is done for morale reasons.  It does not mean that that airplane is assigned to you alone.
Names are painted based on seniority in the squadron.  Typically there are more crews in a squadron than airplanes, so the junior most pilots will have to wait for some time to have their name on a plane.  It is a minor rite of passage when it happens.
For example, in the EA-6B squadrons I served in there were 4 airplanes, and typically 6 crews assigned.  The senior 4 pilots got their names painted, and if memory serves me correctly, the Naval Flight Officers, (NFOs, or WSOs...) who were assigned to their crews were also painted on the same plane.  So, while a junior pilot might need to wait a year or two, a brand new NFO assigned to the Commanding Officer's crew might get on right away.
They are painted on with a stencil whenever a change is called for.  It is low priority in the maintenance scheme of things, so when a senior pilot transfers out and someone moves up the rung it might take several weeks before the plane is moved into the hangar for some other maintenance interval, and the name is repainted then.


text:
How does the modulation of the transponder work?
The transponder works in PCM (pulse code modulation): how does it work in the transponder? I can find explanation of what PCM is, but not if the signal sent from the transponder antenna is digital or analog, and I do not understand which are the advantages of using PCM.

About PCM

I do not understand which are the advantages of using PCM [in a transponder]

Pulse Code Modulation (PCM) is a method used in codecs, while modulation is a method used in modems. PCM with modulation in the name is a misnomer, PCM doesn't relate to modulation in any way.
PCM is used in audio processing to take an analog signal and encode it into a digital stream by sampling, quantization and companding. It is for instance used for audio storage on a Compact Disc and in a Waveform file (.wav). It can be used for transmission, but CPM is only the portion of the chain which prepares the digital data to be sent.
Some forms of PCM are oriented for audio compression (e.g. DPCM or ADPCM), this aspect might have some interest regarding the larger mode S data blocks, but this compression is lossy, meaning some data would not be transmitted, which is indeed not an option for SSR.
To clear any doubt about PCM (and associated encoding delta modulation, delta-sigma, and differential PCM) not being a modulation technique, from Introduction to Analog and Digital Communications:

A point that needs to be stressed one last time: In reality, PCM, DM,
  and DPCM are source-encoding strategies, whose purpose is to convert
  analog signals into digital form. For actual transmission of the
  encoded data over a communication channel, the discrete form of
  pulse-amplitude modulation (PAM) is typically used.

Afaik, there are no PCM application in a transponder or in the secondary surveillance radar to which the transponder replies. It would be interesting to see the document where PCM is mentioned in the context of SSR or modulation.
About transponder modulation

How does the modulation of the transponder work?

Usually a SSR (secondary surveillance radar) ground station or an anti-collision system interrogates aircraft transponders by sending interrogation pulses on 1030 MHz. The transponders answer on 1090 MHz. There are several modes which are used to obtain different information from a compatible transponder. Civil aviation modes are:

Mode A to get the 12-bit identification/squawk code.
Mode C to get the 12-bit identification and the altitude.
Mode S to get 24-bit ICAO identification and various data. In this mode, the transponder also regularly broadcasts data spontaneously for ACAS and/or ADS-B uses.

Modes A/C can be identified by the time elapsed between pulses P1 and P3 (more here) and mode S by the presence of pulse P6 datablock. Transponders answer if they are compatible with the interrogation mode only.
Type of modulation
The basic type of modulation used for everything except the datablock of mode S interrogation (pulse P6) is the On-Off Keying (OOK), a type of Amplitude Shift Keying (ASK) for which amplitude can have two values. Bits of information are transmitted one by one, the carrier is sent or interrupted, according to the bit value:

OOK is a simple class of modulations, used to transmit information at low speed.
Mode S
Mode S (Select) allows the interrogator to specify the ID of the interrogated transponder, to prevent all other transponders to answer. This protocol requires a significantly more information both for the interrogation and the reply. This information is not transmitted as simple pulses using OOK, some form of synchronization must be added, specially when long chain of 0s or 1s are transmitted.
Instead of representing 0 by the absence of carrier and 1 by the presence of carrier (OOK), bits in the P6 datablock are represented by a presence of carrier which some characteristic is changed:

For interrogation, pulse P6 contains information for the Synchronization Phase Reversal (SPR) signal, the interrogation command, the target transponder ID, and 24 bits of the cyclic redundancy check (CRC).

P6 bits are sent using a variant of Differential Phase Shift Keying, the Differential Binary Phase Shift Keying (DBPSK) with phase shift being 0° for 0 and 180° for 1, i.e. consisting in inverting the carrier when bit value is 1:

DBPSK used for in interrogation data

The bit rate is 4 Mbps. In addition, the difference of amplitude of pulse P5 (sent in all directions) and pulse SPR (sent only forwards) is sensed to mute transponders not in front of the interrogator antenna.
Reply data blocks are encoded using Pulse Position Modulation (PPM) which consists in sending a pulse of carrier with a time shift dependent on the value. This ensures a return to zero which is a form of synchronization. For binary values, the pulse is sent either in the first half of the time slot for the bit (for bit 1), or the second half (for bit 0).

PPM with only two values is therefore equivalent to first coding data using Manchester code, then modulating the carrier using OOK already used for mode A/C. This choice was done to simplify airborne electronics when mode S was adopted by FAA.

PPM used for reply data

The bit rate is 1 Mbps. The content of the data block depends on the type of interrogation, or the type of automatic broadcast used. Example for ADS-B squitter sent automatically:

Source

More on ATC interrogation and reply formats here.

GPS satellite transmissions do not use on-off keying. They use BPSK (and BOC for some of the modernized signals). This is the case for both the DSSS spreading code and the navigation message overlaid on top of it.

@pericynthion: That's true. As a (really) poor excuse, let's say BPSK is dual-ASK using two superimposed carriers! Fixed. Thanks.

There are two kinds of transponders in use. Air Traffic Control Radio Beacon System (ATCRBS - pronounced at-crabs) is the older system that uses Mode A for ATC code and Mode C for altitude reporting. The newer Mode Select (Mode S) was introduced to support TCAS.  It supports all three modes A, C, and S.
Mode A and C are pulse amplitude modulation (PAM).  Both modes provide a 12 bit reply word. All pulses are transmitted between 2 framing pulses that are 20.3 micro-seconds apart and are nominally 0.45 micro-seconds long. Each "bit" has a location at a fixed time point after the first framing pulse. The presence of a pulse is a "1" and the lack of a pulse is a "0". There's actually room for a 13th bit, but it not currently used.
Mode A encodes the assigned ATC code - 4 numbers of 0 to 7. The 0 to 7 range aligns with 3 binary bits octal encoding. Thus 3 bits represent each of the 4 numbers - 12 bits total.
Mode C uses the 12 bits to encode the altitude in hundreds of feet using Gillham encoding (Gray code).
Mode S replies have much more data. They have a preamble of 4 pulses. Following the preamble there are either 56 or 112 bits of data encoded with pulse position modulation (PPM). In this coding, each 0.45 usec pulse is placed in the first or second half of a 1 usec slot. If the pulse is in the first half of the slot, it is a "1". If it's in the second half, it's a "0".
Mode S replies are quite numerous and contain multiple formats and data fields.

The transponder is sending digital data (baro, id, sq, etc.) and they are encoded into the reply of the transponder. The modulation of the digital signal to RF in Mode S into a block 56 bit or 112 bit block utilizing pulse position modulation (PPM).  Also the data is aggregated with a CRC to assure data integrity of the fields.
The strategies for Mode A and C transponders are notably different from Mode S.  Mode S is a datalink type system, and is designed to specifically address an aircraft.  Mode A and C are not. Commonality however, includes the use of 1030 and 1090 MHz bands.
PCM is used to efficiently code data minimizing digital "bits" and pulse position modulation (PPM) is used to modulate the digital data into an RF signal to transport the digital data to a receiver.  Strategies not utilizing PCM may take more data to accomplish the same effective data rates.
The digital data on that signal is a combination of the data fields for that transponder strategy, which is packaged and CRC added for error detection and tolerance. The digital data is PPM modulated for transmission via an 1090 MHz reply.

Currently there is no phase encoding in replies from the transponder. All transponder signals on 1090 MHz are either pulse amplitude modulated (Mode A/C) or pulse position modulated (Mode S). Phase overlay on the pulses is currently in development.

@DeltaLima, I checked further, and you are correct, and I am modifying my answer accordingly.


text:
How can I calculate maximum rate of climb?
By what way and with which variables could you determine a plane's maximum rate of climb per time? If I'm not mistaken, I'm looking for VY.

Are you asking "How is Vy calculated?", or "How many feet per minute can I get out of a given aircraft in a sustained best-rate (Vy) climb?" (The former is easy to answer. The latter, well, it depends...) -- Or are you asking something else entirely and I'm just being addle-minded?

Start your plane, taxi to the runway, adjust for V_y and look at your vertical speed indicator… ;) probably the best way to actually know the real answer.

To calculate your possible climb speed $v_z$, you will need

Your engine's thrust $T$
Your airplane's drag $D$
Your airplane's mass $m$

Calculate how much power is needed to overcome drag, and any excess can be used for climbing: $$v_z = v\cdot sin\gamma = v\cdot\frac{T-D}{m\cdot g}$$
Note that this equation makes use of several simplifications, but works well for propeller and slow turbofan aircraft with moderate flight path angles $\gamma$.
To do this with more precision, you need to account for the fact that the aircraft should accelerate during the climb to stay at the same polar point. Now you further need:

The gradient of air temperature over altitude (lapse rate $\Gamma$)
The local speed of sound $a$, and
The gas constant $R$ of air.

You need to add a correction factor $C$ which has several components: $$C = 1 + \frac{1}{2}\cdot\kappa\cdot R_w\cdot\Gamma_w\cdot Ma^2 + \frac{(1+0.2\cdot Ma^2)^{\frac{\kappa}{\kappa-1}}-1}{(1+0.2\cdot Ma^2)^{\frac{1}{\kappa-1}}}$$
where $\kappa$ is the ratio of the specific heats of air and is 1.405, the index w denotes the wet adiabatic gas constant and lapse rate of air, and $Ma$ is your flight Mach number. $\Gamma$ can vary between -0.004°/m and -0.0097°/m, but if you use the average of -0.0065°/m, this equation can be simplified to: $$C = 1 - 0.13335\cdot Ma^2 + \frac{(1+0.2\cdot Ma^2)^{3.5}-1}{(1+0.2\cdot Ma^2)^{2.5}}$$
which is probably the form you will find in most books which care to cover this topic. The second summand takes care of the reduction of atmospheric temperature with altitude and disappears in the stratosphere, and the third summand covers the additional energy needed for acceleration in terms of flight Mach number.

Acceleration factor over the flight Mach number in the troposphere for the Standard Atmosphere
Now your climb speed becomes $$v_z = \frac{v}{C}\cdot sin\gamma = \frac{v}{C}\cdot\frac{T-D}{m\cdot g}$$
As you can see from the graph above, the correction factor is only important at higher speeds, but will cut the climb speed in half at Mach 2. Some jet aircraft need to climb at a high constant Mach number, and then the aircraft needs to decelerate while climbing. Now the correction factor becomes smaller than unity and the climb speed gets a boost because kinetic energy is converted into potential energy while climbing.
Optimum speeds
To pick the flight speed where the climb speed or the fight path angle reaches a maximum, you now need to describe how thrust will change with flight speed. To simplify things, we can say that thrust changes over speed in proportion to the expression $v^{n_v}$ where $n_v$ is a constant which depends on engine type. Piston aircraft have constant power output, and thrust is inverse with speed over the speed range of acceptable propeller efficiencies, hence $n_v$ becomes -1 for piston aircraft. Turboprops make some use of ram pressure, so they profit a little from flying faster, but not much. Their $n_v$ is -0.8 to -0.6. Turbofans are better in utilizing ram pressure, and their $n_v$ is -0.5 to -0.2. The higher the bypass ratio, the more negative their $n_v$ becomes. Jets (think J-79 or even the old Jumo-004) have approximately constant thrust over speed, at least in subsonic flow. Their $n_v$ is around 0. Positive values of $n_v$ can be found with ramjets - they develop more thrust the faster they move through the air.
The flight speed for maximum rate of climb ($v_y$) is reached at a lift coefficient $c_L$ of $$c_L = -\frac{n_v+1}{2}\cdot\frac{T\cdot\pi\cdot AR\cdot\epsilon}{m\cdot g}\cdot \sqrt{\frac{(n_v+1)^2}{4}\cdot\left(\frac{T\cdot\pi\cdot AR\cdot\epsilon}{m\cdot g}\right)^2 + 3\cdot c_{D0}\cdot\pi\cdot AR\cdot\epsilon}$$
which can be simplified for piston-powered airplanes with constant speed propellers as
$$c_L = \sqrt{3\cdot c_{D0}\cdot\pi\cdot AR\cdot\epsilon}$$
The optimum flight speed for this specific case becomes
$$v_y = \sqrt{\frac{2\cdot m\cdot g}{\rho\cdot S\cdot\sqrt{3\cdot c_{D0}\cdot\pi\cdot AR\cdot\epsilon}}}$$
whereas the steepest climb is possible with a $c_L$ of $$c_L = -\frac{n_v}{4}\cdot\frac{T\cdot\pi\cdot AR\cdot\epsilon}{m\cdot g}\cdot \sqrt{\frac{n_v^2}{16}\cdot\left(\frac{T\cdot\pi\cdot AR\cdot\epsilon}{m\cdot g}\right)^2 + c_{D0}\cdot\pi\cdot AR\cdot\epsilon}$$
Nomenclature:
$c_L \:\:\:$ lift coefficient
$T \:\:\:\:$  thrust
$m \:\:\:\:$  aircraft mass
$g \:\:\:\:\:$  gravitational acceleration
$\pi \:\:\:\:\:$  3.14159$\dots$
$AR \:\:$  aspect ratio of the wing
$\epsilon \:\:\:\:\:$  the wing's Oswald factor
$c_{D0} \:$  zero-lift drag coefficient
$\rho \:\:\:\:$  air density
$S \:\:\:\:$  reference area

I have no idea if these formulae are correct, but an excellent answer :p

@JonStory: They are approximations, but are very good when the drag polar of the aircraft is parabolic, as in $c_D = c_{D0} + \frac{c_L^2}{\pi\cdot AR\cdot\epsilon}$

haha I wasn't questioning the equations, just stating that my upvote takes no account of the fact that the maths went way over my head. The description and justification got my upvote :)

@JonStory: I think your skepticism is quite appropriate. Don't trust any equation if you do not know the simplifications and assumptions that were used for deriving them! Maybe you are too generous with upvotes ;)

@PhilCrowther You are absolutely correct, the n$_v$ terms will become zero. But there is a sum under the square root, so a bit remains. I added another line to show what remains in case of n$_v$ = -1.

@PeterKampf And, therefore, the climb speed V(fps) = sqrt(W / (.5 * p * S * cL)) where W = weight(lbs), p = air density (slugs), S = wing area (ft^2) or is that too simplistic?

@PhilCrowther Yes, a bit too simplistic. Pounds is mass, not force (unless you mean pounds times 32.174 ft/s² which gives pound-force). This will become obvious when using SI units: Best climb speed v$_y$ needs mass times gravitational acceleration in the numerator. See my new edit for details.

@PeterKämpf Yes, I was using W = m*g.

@PeterKämpf It appears that the cL equation for piston-powered airplanes with constant speed propellers solves for cL at the point where induced drag is 3X the parasitic drag. The next equation above gives the corresponding speed - which appears to be the level flight speed in which Lift = m X g. However, equations relating to aircraft in a climb indicate that the Lift should equal m X g X cos(angle of climb). In practice, should the cL computed above be reduced to reflect the estimated angle of climb?

@PhilCrowther Yes, to be precise, c$_L$ should be reduced. But these are approximations for small climb angles. Using a quadratic polar is already an approximation and assuming lift to be parallel to weight is another one. Still, the result is useful even though a precise calculation would look a lot more complicated and have a slightly different result.

Variables will have to include air conditions (temperature, humidity, altitude starting from), engine performance data at said altitude, and the vy speed for the airplane in question.

You are correct that Vy will give you the max RoC. However Vy is actually only a speed, not a climb rate, and will correspond to slightly different rates of climb depending on a few factors. The ones that spring to mind are A/C weight, temperature, and density altitude. Along with power setting but that one is somewhat self explanatory. As for determining the speed, normally you would just consult your PPOH or AOM to find a performance chart that will give you the base speed and you can correct it for each of the variables listed in order to have a precise climb rate.

It's important to note that reference speeds like Vx and Vy are not constants for a specific aircraft. They vary with air density and airplane weight. Many student pilots wrongly assume it's a fixed airspeed number. In some checklists, the pre-takeoff checklist mentions setting airspeed bugs for Vx, Vy and best glide based on current conditions and gross weight.

Also note, that Vy is only a sustained climb, if you have enough speed to bleed off, I'm betting you can get a better RoC by pointing the nose skyward, albeit not for very long.

Yes, that's how we start our zero-G fun :-)

@roe Indeed, and if you've got excess airspeed and a rapidly approaching line of trees carefully executing that tradeoff of airspeed for altitude in the climb may be the difference between clearing the tops and having to climb the tree later to retrieve your tires. (Not that you should get yourself into such a situation to begin with, but we've all seen youtube videos...)


text:
Is Aeroflot flying airliners without brakes?
I have just read the news that the Russian flag carrier Aeroflot has deactivated the brakes on nine of its aircraft. Not to dismiss the journalistic talent behind these reports, but could it be that this is one of those cases where a journalist picks one issue and blows it out of proportion to create a shocking story? Alternatively, is this a somewhat viable approach to fly when spare parts for the aircraft are not easily available?

@RTO: maybe in Putin's Russia you cannot say no to some requests.

The article has this update: "Aerotime apparently translated the Moscow Times story and some associated memos from Aeroflot and reported the planes are allowed to fly with the deactivated brakes for up to 10 days, but it’s not clear what will happen when the time is up. It’s an accepted practise to deactivate the brakes on one or more wheels on multi-wheel trucks but it’s not clear the extent to which Aerflot is doing it"

Can you link to where you read this news please ?

This is just a wild misinterpretation / mistranslation.

The Aeroflot warning, on which the article is based, states:

The exact translation of the message's title is:
"Safety culture (braking with a deactivated brake)". With this wording, "brake" can not be a substitute for "brake or brakes", it would've been explicit.
The message says "Our crew has asked to warn their colleagues about flying with a deactivated brake. When braking, the plane might tend away from the runway's centerline. You should address this during the approach briefing. Pay special attention with wet runways and side winds! The aircraft will tend to turn towards the side where the brakes have not been deactivated."
This last phrase puts a nail in the coffin for the interpretation that there are no brakes - it specifically states that there is a side where none have been deactivated.
It further lists aircraft types which have deactivated brakes, which includes 5 B777s. According to the Boeing 777 Minimum Equipment List, section 32 (page 236), -45-01A:

(M)(O) One per each six-wheel truck
may be inoperative provided:
a) Associated brake is
deactivated with a
deactivation assembly, and
b) AFM performance
decrements for brakes
deactivated are applied.

While the advice seems obvious, it's possible that some pilots have expected the plane to automatically match the braking force on both sides, compensating for a deactivated brake, and were caught off guard when it didn't.
The telegram group reacts to the message as obvious, until another airline's pilot says "We had some deactivated brakes too, the MEL permits it. Until we got an anti-skid failure while taking off from a wet runway. Then on CEO's orders we limited the use of this MEL to returning to base only".
It's not clear whether it's connected to a parts shortage. Aeroflot has 178 aircraft and the list has 9 that have an inoperative brake. This is 5%, which is within the norm for a fleet.
However, the figure is strangely higher at 20% for the B777. I'm not sure what the typical occurrence rate of an inop brake is for that airframe - would appreciate any comment from 777 drivers.

Nice find!  Thank you for definitively putting the silliness of "they're flying around with no brakes at all!!!!" to bed.

@RalphJ Yeah, landing a 777 with no brakes would probably require a runway 10 or more miles long. And good luck taxiing it.

The very presence of this notification sort of implies that the condition is much more common than it used to be, and more pilots encounter it in practice. While inactivation of a brake is not so unusual and all pilots must be aware of this possibility, it seems that the prevalence of it in normal times must be much less than 5%. I've seen reports that the situation can be bad enough at times that airlines (not necessarily Aeroflot) must get "creative" and move the inop brake between aircraft, gaining legal time extension every time.

@Zeus It's possible, but there's not enough evidence to be sure either way. You rarely hear about the memos that don't make it into a news article. Except for the B777, the prevalence is as expected. I hope someone who flies, maintains or manages 777s chips in - do they generally go through more brakes than most, or is that unusual.

It is most certainly not legal* or safe for an aircraft to operate with all its brakes disabled. News reports indicate that the aircraft are 5 Boeing 777 and 4 Airbus of various types. They are less clear on whether some of the brakes are disabled (legal under certain circumstances) or all of the brakes are disabled (which is certainly illegal).
A quick glance at the B777 Master Minimum Equipment List (which can be found here using the FAA's search engine) indicates that "One [brake] per each six-wheel truck may be inoperative" given specific conditions. Having two or more brakes in one truck inoperative makes the plane unairworthy. Furthermore, the MMEL does not permit an operator to leave parts permanently inoperative- the brakes are in repair category "C", indicating that the issue is required to be rectified within 10 calendar days.
It's possible that the airliners are being operating legally with the minimum number of brakes required to be airworthy, and that the pilots are simply requested not to use the remaining brakes to prevent wear on the remaining operative ones.
There is a message, reportedly from Aeroflot to its employees, that can be found here. Google translate indicates that its a warning that when the brakes are applied with the brakes deactivated, the plane may pull to one side. Obviously, this doesn't make sense if all the brakes are deactivated! That is a point in favor of the theory that some of the brakes are disabled, but not all of the brakes.
*Under ICAO regulations, anyway. Russia can set whatever regulations it likes within its airspace.

@RTO I suspect (and hope!) something was lost in translation and they mean "less than all the brakes," not "no brakes."

You're looking at the FAA certification. Russia is not subject to the FAA; they have their own certification regime, the IAC/MAK, which does not have to follow US rules. In fact, European EASA rules differ from the FAA's. For example, there are 737 classics flying and registered in Europe which would be illegal for US registry because they lack fuel tank inerting.

@user71659 The MMEL is made by the manufacturer and then approved by a national authority. Sure, there might be minor differences, but I don't think anyone seriously thinks that Boeing would produce an MMEL that allows for all the brakes to be inoperative.

@Chris That argument is nonsensical: if a plane operated without brakes it would end up in a ditch the first taxiway intersection it comes to. The FAA MMEL for a 737 says you have to fix the fuel tank inerting in 10 days due to the risk of a fuel tank explosion. EASA planes don't even have to have one installed. The fact that IAC-MAK/CIS planes can have significantly different certification standards is absolutely nothing special.

@user71659 "If a plane operated without brakes it would end up in a ditch the first taxiway intersection it comes to." So we're in agreement that there is absolutely no way that any form of the MMEL that Boeing would produce would allow the 777 to be operated with all brakes disabled, right?

@Chris All brakes inop is absurd. However, the linked article doesn't claim that. It claims some brakes inop, which you state is valid to a certain degree under FAA MMEL. The question here is to see what the IAC-MAK MMEL says. For all you know, it could say no brakes inop because it snows in Russia.

@user71659 For that matter, it's also possible that the MEL for the individual planes doesn't allow any brakes to be inop even if the IAC-MAK MMEL says it's okay. I find both of those things unlikely. It's not like Russia is all snow 24/7/365. I only use the FAA MMEL to show that Boeing considers some brakes being inop to be acceptable, and so it's possible that the Aeroflot aircraft are being operated legally. If you know a more directly applicable document that is available online (the IAC-MAK MMEL or even the Aeroflot MEL for one of the planes), I'm more than happy to replace the link.

This message looks like useless bureaucratic garbage. "We are obliged to inform that braking without all brakes can be different than braking with all brakes". In addition, mesasge to which this image was attached ( https://t.me/aviatorshina_chat/166719 ) basically repeats this answer: you can fly up to 10 days with some brakes deactivated as long as specific conditions are met.

There's no reason to believe this to be untrue, it is reported in the Moscow Times which is a Russian language web publication - somewhat ruling out "foreign" propaganda.

So, as of  July 31, nine Aeroflot aircraft were flying with deactivated brakes. Among them are five Boeing 777s, two Airbus A321s and one A320 each with an A330.
Airlines resort to such measures when it is not possible to immediately replace the brake due to a malfunction or wear of the disc to the permitted limits. At the same time, it is allowed to fly with a refusal for no more than ten days, according to safety instructions.

Note that it does not say all brakes, it is quite possible they're operating without some of the brakes completely within the manufacturers recommendations for when a part is (temporarily) unavailable.
This is all part of an ongoing problem where sanctions due to the war in Ukraine have severely limited Russia's ability to import spare parts for their western-built fleets.
The only viable alternative, and one which it has been practicing for well over a year now is to cannibalize it's fleet to keep a certain number flying safely.

@Jamiec The Moscow Times is banned in Russia and headquartered in the Netherlands, I'm sure it can be called "foreign"

I suspect the quote makes better sense if one infers that "flying with deactivated brakes" actually means "flying with {a small but non-zero number of wheels with} deactivated brakes" and not "flying with {all brakes on all wheels} deactivated". My first take on the quote was the latter, yielding the reaction that the story was clearly nonsense.  On reflection, I could see the former understanding being consistent with the text too.

You mean a journalist got a detail about aviation wrong?  Possibly not the first time...  Or perhaps they deactivated the autobrakes on some number of aircraft (not the anti-skid -- that's dry-runway only and big performance penalties for a 737, and probably similar or worse for bigger jets), and the difference between "brakes" and "autobrakes" got lost during the journalism.


text:
Are fighter jets designed to be so inherently unstable that a human can't fly one unassisted?
I'm told that in order to be more maneuverable fighter jets are designed in a way that makes them impossible for a human to control without the help of a flight computer.  Is this actually true?  Would a modern fighter (like an F-22 or Su-35) crash if the stability computers died?
Further, if this is true, what design techniques are making them unstable and how do they help with maneuverability?

My short answer:

Stability is reduced by shifting the center of gravity aft.
Shifting it past the neutral point makes the airplane unstable, so movements away from the trimmed state are accelerated. This increases maneuverability.
Flight computers are multiple redundant, if one dies the others take over.
Slow unstable airplanes can be flown by a human pilot, but not fast unstable airplanes.

For the long answer, let me first clarify terms:
Static stability is the tendency of a system to return to it's old state after being disturbed. Take a pendulum: If you pull it to one side, it will return to the middle. Eventually.
Dynamic stability is the tendency of an oscillating system for the oscillations to die down over time. Take the same pendulum: It will swing from side to side, and  friction will ensure this happens with ever smaller amplitude.
Now we need to add dimensions, all three of them: Pitch, roll and yaw. An airplane can be stable in one dimension and unstable in a different one. I understand your question such that you ask about the static pitch stability (or longitudinal stability) of fighter aircraft.
The Wright Flyer was longitudinally unstable (see here for more). Once aircraft designers learned that aircraft can be made to fly stable, and learned that this is of immense benefit in pilot training, static stability became a requirement for new aircraft. When war in Europe broke out, the British forces were equipped with a superb training aircraft, but it was so stable that it took effort and time to convince it to change course. They were shot down in droves.
From now on, low stability was a prime requirement for fighters and aerobatic aircraft. Static stability is proportional to the control forces (more precisely: To the hinge moment of the respective control surface), so reducing stability gave pilots more response for the same effort. Longitudinal static stability is measured as the relative distance between neutral point (NP) and the center of gravity (CG). See here for more. Longitudinal static stability is achieved by placing the CG ahead of the NP. Shifting the CG back gives you a more responsive airplane, but also one which is more easily disturbed by gusts.
This is the design technique you asked about. Pretty simple, right?
Once you shift the CG aft of the NP, stability is lost and the airplane will increase deviations from the trimmed state. This can be helpful if you want large angle changes, and quickly. An unstable aircraft only needs a small kick and will do the rest of the maneuver all by itself.
This is how it helps in maneuverability. But it is even more helpful to reduce the inertias, especially around the roll axis, for a faster response. That is why all combat aircraft have their engines close to the center.
Of course, negative stability is not acceptable when you need to take your hands off the stick to get a map out or to pee on a long flight. So without computer control, the limit was a CG position near, but not aft of the NP.
With supersonic aircraft, things got more complicated. Now the aircraft operates in two flight regimes, one where lift acts at the quarter chord of the wing and one where it acts at mid-chord. Aircraft with low static stability become very stable in supersonic flight, and the tail surface has to create a high down-force so that the sum of all lift stays where the CG is. Creating lift always incurs a drag penalty, and in supersonic flight it has to be payed twice: One for the excess lift on the wing (which is needed to compensate for the tail's down-force) and one for the down-force on the tail.
Using a flight control computer offers the possibility to allow the pilot to let go of the stick without the aircraft going off course. Now the stick does not command elevator deflection, but pitch rate, and the CG can be moved back from maybe 12% of MAC (mean aerodynamic chord) to -2%. If you compare the wing areas of stable and unstable jets (Jaguar and Mirage F-1 are prime examples), you will see how much is achieved just by going back with the CG by a few percent of wing chord.

Conventional SEPECAT Jaguar and CCV version (picture from Ray Whitford's Fundamentals of Fighter Design). Both configurations have the same airfield and combat performance!
Can a human still fly such an aircraft? In glider competitions, the more daring pilots fly with relaxed static stability and have no problem to keep the aircraft under control. Even the Wright brothers could handle their unstable airplane, and handling improved when they moved the CG further back (If you want to know why, please post a new question. This answer is getting too long already!). However, the speed of an aircraft's pitch response is proportional to flight speed (and inverse to the pitch moment of inertia), so faster aircraft are harder to control. You can compare dynamic pressure with the stiffness of a spring: A stiffer spring moves the eigenfrequency of a spring-mass system up, and the same is true for the eigenvalues of the equations of motion of an aircraft. Given that the reaction time of a good pilot is at least 0.1 s (and more if he/she is tired), it is impossible to counteract motions with frequencies of more than a few Hertz. The lag means that the reaction comes too late and will support the motion. See this YouTube clip how that works out in practice. This crash was due to wrong signal gains, not a classic instability (after all, the flight computer was still working, but produced too strong elevator deflections).
I would venture to say that a human can still barely fly an unstable jet at low speed (after all, Tom Morgenfeld almost got the YF-22 under control), but once he firewalls the throttle, he will be always behind the plane, and will crash it soon.
Size helps: Bigger planes have lower eigenfrequencies, and light, big vehicles are easy to control, regardless of stability. All Zeppelins were completely unstable in yaw and above the critical airspeed of an airship (again, please ask for a more detailed answer on this aspect) also in pitch, but with one person each for the vertical and the horizontal control surfaces, and enough people on board to rotate them after 2 - 4 hours, nobody felt the need to make Zeppelins naturally stable.
If one computer dies, the others take over. Most unstable configurations have four parallel computers which cross-check their result to catch any malfunction. The Dassault Rafale uses only three, but adds safety by clever algorithms for checking the results.

Excellent answer!  I wonder, though, if you could possible put a one paragraph tl;dr version at the top of the post?  For people who aren't quite as enthusiastic for knowledge as I am?

Even after four months nobody was curious enough to ask about the critical speed of airships. I guess my answers are really too long.

@PeterKämpf i am curious to read about it. Why not make a selfie question and you can also get the hat!

If the flight computers did all fail, would the plane immediately go into a flat spin, head over tail spin, or would it vibrate itself to pieces? What is the failure mode?

I agree with vasin1987, your knowledge is impressive...share!

@RoboKaren: It would either pull up or tuck, depending on the initial disturbance. Depending on flight speed and altitude, it could stall, disintegrate or become a lawn dart. Flat spins are one possible outcome, but not likely - they need some help from the configuration (long fuselage, oval nose shape with vertical long axis, obstructed airflow at the fin at high angle of attack).

Instability should actually increase at slower speeds.  While it is true that smaller movements create more dramatic effects at higher speeds, slower speeds require much larger control inputs (at some point I imagine full throws to maintain balanced flight), and it would be many times easier to induce PIO in an unstable, unassisted platform.  The Super Hornet is a testament to slow speed maneuverability via computer.

@RoboKaren in the case of the Super Hornet the aircraft becomes an 80 million dollar lawn dart.  The control surfaces won't even move if you lose all of the computers.  Modern fighter aircraft have advanced beyond fly-by-wire into the realm of control-by-wire, whereby the computer actually decides what to do based on what it thinks you're trying to accomplish with the given inputs.

A useful related point may be that very lightweight slow-flying aircraft such as human-powered aircraft, some ultralights, etc may experience so much pitch damping (due to radius of curvature of flight path being so small) that they may be controllable even if statically unstable in pitch.  First encountered this in a textbook; may or may not have been "Stability and Control" by Abzug and Larrabee.  May also play a role in ideas in "Circular airflow and model aircraft" by Frank Zaic.

Yes, the first such airplane was the F-16.  It was designed as inherently aerodynamically unstable, which allows it to respond superbly in combat.  This was made possible in that it is a fly-by-wire aircraft.  Maneuverability is increased, because by definition it is the ability to change states.  Stability is the resistance to change.  The more stable you are, the harder it is to turn/pitch quickly in a dynamic situation.
And yes, a pilot would not be able to land these aircraft if the fly-by-wire systems became inoperative.  There are instances where F-16 pilots have lost their computer and have died because of Pilot Induced Oscillations - the condition where the pilot isn't correcting for their aircraft's instability at a rate fast enough to maintain control.
Other such unstable aircraft are the B-2, F-22, F-35, Eurofighter, etc.  All modern fighters need to be inherently unstable to be competitive.

actually you have "Pilot Induced Oscillations" when the pilot is just barely fast enough to keep up with the movement of the aircraft, but due to the delay into the brain-muscle reaction it is out of phase (i.e., he does the right thing at the wrong moment).

Actually the principle design driver for the unstable configuration is for supersonic performance as an unstable design(subsonically) reduces the supersonic trim drag, it has little to do with manoeuvrability and there are arguments that it is detrimental. From a pilots perspective he wants to point the nose, the more unstable the design the harder it is to stop the nose at the required attitude. Finally this is NOT an example of a PIO, the clue is in the name, the oscillation is NOT induced by the Pilot, if the pilot let go of the stick the oscillations would not subside. ergo not PIO.

RE Federico - thanks for the catch on the typo, corrected.  I accept your explanation for PIO, but was trying to simplify the explanation.

RE Adrian: PIO first - the oscillation is pilot induced, but not the initial deviation.  The oscillation doesn't come from the unstable design, but from the pilot trying to correct a deviation from stability.  Still PIO.
Your statements regarding supersonic stability are also true, but I wouldn't say that it is more of a driver than maneuverability.

If the system is unstable and the computers fails then the aircraft will depart whether there is a pilot in the loop or not. In a PIO it is the feedback loop around the pilot that unstable, so if the pilot can go open loop, i.e. he physically lets go of the stick then the aircraft will return to a stable state. For highly unstable aircraft if the FCS has failed then the aircraft will depart irrespective of what the pilot does, which is why this is not a PIO. It's also irrelevant because if the FCS has failed then it won't respond to any pilot input so the pilot can't induce an oscillation.

Instability in pitch lowers trim drag for an aircraft with a tail.  Stability refers to the relative positions of the center of lift (cl) and center of gravity, (cg).  When the cg is ahead of the cl (stable) an aircraft that stalls can fall forward, increasing speed and recovering.  When cl is ahead of cg this will not happen.  However, when cl is ahead of cg the tailplane produces lift.  When cg is ahead of cl (unstable), the tailplane produces a downward force, lowering efficiency.  For a canard, stability lets both surfaces produce lift.  However, the canard fighters tend to be unstable in order to increase pitch rate and lower supersonic drag.  At supersonic speeds cl moves backwards meaning the canard would have to shoulder a larger part of the burden of lifting the aircraft.  Since the canard is a less efficient lifting surface than the wing, this is undesirable.
Edit:
I'll add that as a fighter goes supersonic it goes from unstable to stable for both canards and tails.

A stable pitch axis means that when the aircraft is disturbed, e.g. a gust then the resultant pitching moment counter-acts the disturbance, stall is more a loss of lift due to flow breakdown over the wing.

The center of lift is always at the same lengthwise point as the center of gravity if the aircraft is trimmed in pitch. Probably you mean the center of lift of the wing-fuselage combination. Better yet, use the term neutral point. Also, the aircraft is longitudinally stable when the cg is ahead of the neutral point. You seem to think it is the other way around.